A hemispherical network of radius a is made by using a conducting wire of resistance per unit length ‘r’. Find the equivalent resistance across OP

To determine the equivalent resistance across points O and P in a hemispherical network made from a conducting wire, we'll first understand the structure of the network and how the resistances are arranged. The hemispherical shape means we have a surface that is a half-sphere, and the wire resistance is uniform across its length. Let's break it down step-by-step.

Understanding the Geometry

Imagine a hemisphere with radius 'a'. The wire forms the surface of this hemisphere. When we look at it from above, we can visualize the points O and P being at the two ends of the diameter of this hemisphere. The key here is to realize that the current will spread out symmetrically from point O to P.

Resistance Along the Path

The total length of the wire forming the hemisphere can be calculated using the formula for the surface area of a hemisphere, which is \(2\pi a^2\). However, since we are interested in the resistance, we only need to consider the wire's path along the hemisphere.

For a wire of resistance per unit length 'r', the total resistance \(R_{total}\) along the entire surface of the hemisphere is given by:

R_total = r \times L

Where \(L\) is the length of the wire forming the hemisphere, which is \(L = 2\pi a\) (the circumference of the great circle at the base of the hemisphere).

Calculating the Total Resistance

Now substituting for \(L\):

R_total = r \times 2\pi a

Symmetry and Division of Resistance

When the current enters through point O and exits through point P, it will divide symmetrically into two paths (one path going to the left and the other to the right around the hemisphere). Each of these paths covers a quarter of the hemisphere's circumference, which is \( \frac{1}{2} \times L = \pi a \).

The resistance for one path from O to P is:

R_{single} = r \times \frac{L}{2} = r \times \pi a

Combining the Paths

Since there are two identical paths (left and right), we can treat them as resistors in parallel. The formula for two resistors \(R_1\) and \(R_2\) in parallel is:

R_{eq} = \frac{R_1 \cdot R_2}{R_1 + R_2}

In our case, \(R_1 = R_2 = R_{single} = r \times \pi a\). Therefore, the equivalent resistance \(R_{eq}\) across points O and P is:

R_{eq} = \frac{(r \times \pi a) \cdot (r \times \pi a)}{(r \times \pi a) + (r \times \pi a)} = \frac{(r \times \pi a)^2}{2(r \times \pi a)} = \frac{r \times \pi a}{2}

Final Result

In conclusion, the equivalent resistance across points O and P in the hemispherical network is:

R_{eq} = \frac{r \pi a}{2}

This result illustrates how symmetry in the configuration of the network allows us to simplify the calculations significantly. By understanding the distribution of current and the arrangement of resistors, we can effectively analyze more complex circuits.